As you have a "nice form" differential equation here, you can use Separation of Variables and integration to solve. Note that
$$\frac{dP}{dt}=0.4873P^2$$
can be written as
$$\frac{dP}{P^2}=0.4873dt$$
Now you can integrate both sides
$$\int\frac{dP}{P^2}=\int0.4873dt$$ Thus
$$-\frac{1}{P}+C_P=0.4873t+C_t$$
To finish, note that $C_P$ and $C_t$ are arbitrary constants due to the indefinite integrals. Thus we can rewrite as
$$-\frac{1}{P}=0.4873t+C$$
which is the same as
$$P=\frac{1}{C-0.4873t}$$
for some real $C$. Now as $P(0)=0.2, C=5$ So finally we have
$$P=\frac{1}{5-0.4873t}$$
As the denominator approaches $0$, the population approaches $\infty$. This occurs when $5-0.4873t=0$ Thus $t=10.2606197....$ and as $t$ is the number of centuries after $1000$ AD, we can expect an infinite population in $10$ years from now.